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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    When a certain photosensitive surface is illuminated with monochromatic light of frequency u, the stopping potential for the photo current is\[\frac{-{{V}_{0}}}{2}.\]When the surface is illuminated by monochromatic light of frequency\[\frac{\upsilon }{2},\]the stopping potential is\[-{{V}_{0}}.\] The threshold frequency for photoelectric emission is [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) \[\frac{4\upsilon }{3}\]                                       

    B) \[2\upsilon \]

    C) \[\frac{5\upsilon }{3}\]                           

    D)   \[\frac{3\upsilon }{2}\]

    Correct Answer: D

    Solution :

    Einstein?s photoelectric equation in the two cases is given by \[\frac{e{{V}_{0}}}{2}=h\upsilon -h{{\upsilon }_{0}}\]                                       ...(i) and\[e{{V}_{0}}=\frac{h\upsilon }{2}-h{{\upsilon }_{0}}\]                                 ?(ii) From eqn. (i) and (ii), \[\frac{1}{2}=\frac{h\upsilon -h{{\upsilon }_{0}}}{h\upsilon /2-h{{\upsilon }_{0}}}\Rightarrow {{\upsilon }_{0}}=\frac{3}{2}\upsilon \]


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